Order-two intrinsic variograms are conditionally negative definite

Suppose first that $X$ is wide stationary. For every $s,h$, its increments have mean zero because the field mean is constant. Wide stationarity gives $C(h)=\mathrm{Cov}(X(s),X(s+h))$ and $C(-h)=\mathrm{Cov}(X(s+h),X(s))$; covariance symmetry [F3] therefore gives $C(-h)=C(h)$. Both level variances equal $C(0)$, and the cross-covariance in the increment variance is $C(-h)=C(h)$. Thus $\mathbb{E}[(X(s+h)-X(s){)}^2]=\mathrm{Var}(X(s+h)-X(s))=2C(0)-2C(h)\mathrm{.}$ Dividing by two gives $\gamma (h)=C(0)-C(h)$.

For arbitrary sites $s_1,\dots ,s_n$, set $Y_i:=X(s_i)-X(0)$. Each $Y_i$ is square-integrable by order-two intrinsic stationarity, since it is an increment from $0$ to $s_i$. Moreover, $Y_i-Y_j=X(s_i)-X(s_j)$, so [F1] gives $2\gamma (s_i-s_j)=\mathbb{E}[(Y_i-Y_j{)}^2]\mathrm{.}$ All cross products are integrable because $2\mathrm{\mid }{Y}_i{Y}_j\mathrm{\mid }\le Y_i^2+Y_j^2$.

Expanding the finite sum from step 1.2 and using ${\sum }_i{a}_i=0$ yields $\begin{array}{rl}{\displaystyle \sum _{i,j}{a}_i{a}_j\gamma (s_i-s_j)}& {\displaystyle =\frac{1}{2}\sum _{i,j}{a}_i{a}_j\mathbb{E}[Y_i^2+Y_j^2-2Y_i{Y}_j]}\\ {\displaystyle }& {\displaystyle =-\mathbb{E}\text{\hspace{0.17em}⁣}\left[{\left(\sum _i{a}_i{Y}_i\right)}^2\right]\le 0.}\end{array}$ The two sums containing $Y_i^2$ or $Y_j^2$ vanish because each contains the factor ${\sum }_i{a}_i$; the remaining term is the negative expectation of a square. No choice principle is used.